help-38

[drifting rose]

well can you expand on the original question then?

[onyx forge]

Is the answer

Two?

In this

[drifting rose]

no

well yes

[onyx forge]

are you sure its not?

[drifting rose]

no it is, I forgot I didn't give you the whole question

[onyx forge]

Because, what they basically had to do was

make the bottom 2x^2

Such that theota on denominator and numerator is same

Which become 1 basically

[drifting rose]

no I get that

[onyx forge]

uhuh

do u not get why that happens?

[drifting rose]

no I'm confused on why the equated number so it balances out is 2*x^2/x^2

I would assume it would just be like x^2/x^2

why is there the 2 on top?

shouldn't they be the same?

[onyx forge]

AH

[drifting rose]

like if you want to add 1 to like 3x you need to also put a minus 1

[onyx forge]

The original equation was

sin(2x^2)/x^2

[drifting rose]

mhm

[onyx forge]

To get two in the bottom

They should have multiplied with

2/2

Now why they did the stuff on the RHS, no idea

[onyx forge]

But basically all u had to do was this

[drifting rose]

2/2

[onyx forge]

yes

[drifting rose]

wouldn't that cause the 2 to be on the top also

[onyx forge]

yes

since u cant just introduce in a new number

in the denominator

u gotta multiply the top with the same number

so it essentially cancels out

[drifting rose]

I need to know how this works out I am so lost

[onyx forge]

sigh

See

[drifting rose]

it doesn't make sense to me

[onyx forge]

The equation on the right says

When x approaches 0, sin(x)/x becomes 1

If you take sin(0.0001)/0.0001, the answer would be 1

[drifting rose]

don't explain limits to me

we are going of course now lol

[onyx forge]

damn

but, thats the theorem

basically

[drifting rose]

I get the theorem

[onyx forge]

Now, in the question

you have

[drifting rose]

I don't get the equating bit

[onyx forge]

sin(2x^2)/x^2

Acc to the theorem, the number in bracket and denominator must be same

[drifting rose]

I GET THE THEOREM

[onyx forge]

You cant change anything in the bracket, but u can change the denominator

OH GOD

THEN WAHT DO U NOT GET

IM EXPLAINING U TO STEPS TO GETTING THE ANSWER

IN THE QUESTION ??!?!?!?!?

[drifting rose]

I just don't get how the 2 on the bottom equates to 2*x^2/x^2

[onyx forge]

huh

[drifting rose]

becuse if you want 3x with a 1

it'll be 3x +1 -1

because we can't just add random number all willy nilly

[onyx forge]

mate

don't look at the working

solve it yourself

that's what i suggest

I got no idea what yo working has been done but

Do it on your own, solve it, get yo answer and ull be all merry and happy

Cant help u with deciphering that maths

[drifting rose]

so you can't tell me why adding a 2 on the bottom needs an equivalent of 2*x^2/x^2

[onyx forge]

it doesn't need an equivalent of 2x^2/x^2

it just needs 2

Because 2x^2/x^2 is literally 2

[drifting rose]

so why wasn't it just 2, why did they have to add the x^2 to it

[onyx forge]

huh

[sacred sapphire]

they might've just thought just replacing the x² with 2x² is easier to see

since you added the 2x² on the bottom, you must multiply it by 2x² on the top

and the x² before stays

thus 2x²/x²

[drifting rose]

so if I were to multiply them together I would get the original right?

[sacred sapphire]

but its the same as just multiplying and dividing by 2 really

yes

[onyx forge]

...

im gonna cry

[onyx forge]

im so gonna sob

[drifting rose]

oh wait they do cancel

dam

[onyx forge]

BRO

I SWEAR

[drifting rose]

all apart of the process

welp thanks for the help

[onyx forge]

sigh

thanks shioshi

[sacred sapphire]

but yeah this guy was trying to tell you that 2x² / x² is 2

and you put that 2 below the original

which all boils down to divide and multiplying 2

[sacred sapphire]

rough life man

[onyx forge]

made me wanna kms

[drifting rose]

I guess it never occured to me that the equivalent of adding stuff to the demoniator is multiplication :V

[sacred sapphire]

division*

adding stuff to the numerator is multiplication

[drifting rose]

I mean't it in the equating context but yea thats right

alright thanks for the help

soz :V

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[frigid lark]

,tex -sin^{7}(\frac{1}{x})+sin^{7}(\frac{1}{e})

[solid kiln]

suds
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[frigid lark]

i did this

i dont know what im doing wrong

, tex bc i used the definite intergral thing of $ \int_{b}^{a}=f(a)-f(b)

[solid kiln]

suds
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[frigid lark]

and i know using the ftoc its basically unchanged but i dont understand why you can't use the definite integral formula with it, because its the same as integrating the thing and differentiating it which would be -sin^7(1/x) + sin^7(1/e)

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[short cave]

For what values of $n$ does $(x+\frac{1}{x})^n$ have a term independent of $x$ when expanded?

[solid kiln]

BOHO

[short cave]

i applied general term to this and i got n = 2k. does that mean any non-negative even integer of n will have a term independent of x? (as long as k is a non-negative integer)

[sly pewter]

prob

[short cave]

also

Find the values of $n$ such that the expansion of $(3x^2+\frac{4}{x})^n$ has a non-zero $x^2$ term.

[solid kiln]

BOHO

[short cave]

im not sure how to do this

after i get a value of n in terms of k

i got n = 3k - 2 and subbed that in to the general coefficient thingy

and i got

$\binom{3k-2}{k} * 3^k * 4^{2k-2}\neq 0$

[solid kiln]

BOHO

[short cave]

idek if what im doing is right

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@short cave Has your question been resolved?

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@short cave Has your question been resolved?

[deep niche]

In the general term, the power of x will be 2(n-r) - r = 2n - 3r.
You want this to be 0, so r = 2n/3.
For what values of n will this be an integer?

[short cave]

ah, I messed up the general term part anyways.

[short cave]

So would n just be any multiple of 3 for that to be an integer?

[short cave]

aight thanks bro

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[sullen stone]

How do i solve the question that is crossed out? (b)

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Please don't occupy multiple help channels.

[sullen stone]

The translation is, "Point P belongs to the end leg of angle alpha, and point Q - to the end leg of angle beta. Calculate sin alpha-sin beta"

Where the coordinates of point P is (-9,3) and point Q, (2,6)

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[torn sun]

Can someone solve this ?

[split chasm]

why close the other channel

[torn sun]

Someone else was typing for a long time

[split chasm]

you laid claim to it first,

had your name on it

[torn sun]

Nah it's ok.

[split chasm]

because its effectively a dead channel now and time gets wasted
anyway
look up adventitous triangle

that also looks snapshotted from MindYourDecisions,

[torn sun]

He did it by constructing a isosceles triangle. Isn't there any other way ?

[split chasm]

trig bash probably

[torn sun]

Can you do it ?

[split chasm]

trig bash? no

[torn sun]

Ok let's wait ig

[split chasm]

not interseted in doing that

[torn sun]

For someone else

[torn sun]

Ok

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[gusty mulch]

does anyone know how to interpret this one

[gusty mulch]

never seen anything like it before, it says the answer is 3/2 but idk how it got there

[wild imp]

it's just the derivative at x=1, to solve it properly first find f(x) at that point, then substitute the values, and evaluate the limit

[gusty mulch]

F(1) = 1

Idk how to solve the derivative properly at this point @wild imp

How would you go about it

[wild imp]

did you find f(x)?

[gusty mulch]

f(1) = 1

[wild imp]

the exact function not the value

in f(x) = ax+b form

[gusty mulch]

At x = 1 right?

Slope is 1 and y intercept is 1

So x + 1

Am I right?

[wild imp]

no

try again, the slope is rise/run

the line goes from y=-1 to 2 for x=0 to 2

also the y-intercept is -1

[gusty mulch]

Rise is 1 and run is 1

[wild imp]

see its 3 and 2

[gusty mulch]

Omg I was calculating from 0 to 1

Sorry

[wild imp]

so can you find f(x) now?

[gusty mulch]

Yes

Thank you

3/2x -1

Is the function

[wild imp]

yes, thats right

[gusty mulch]

Question

[wild imp]

now you just need to evaluate the limit

[gusty mulch]

Y intercept just means when y plane is crossed

At wherever?

Thank u

[wild imp]

y-axis*

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[rugged crag]

x and y have to cross the line with a dot, but if i choose random ones like -4 and -7 it doesnt cross

[astral ore]

u can choose a random x value

and then you have to plug that x value into:

y = -3x - 2 to get the y value

u don't randomly choose an x and y value

u can only choose one value (x or y) randomly

[rugged crag]

aha

so the i choose the next x to be like idk 4, and then i calculate y

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[rugged crag]

how do i get rid of this

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[plush sequoia]

Hello, can someone confirm if width is adjustable in a frequency table

[plush sequoia]

<@&286206848099549185>

[frail yarrow]

1. Dont ping helpers until 15 min has passed
2. I'm not really sure what you're asking

Could you provide some context?

[plush sequoia]

srry

nvm is goods

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[untold parcel]

can someone explain to me how this changed to that ?

[marble wharf]

they divided by e^b on both sides

or equivalently, they multiplied by e^(-b) on both sides

[untold parcel]

oh wait

nvm

thank you so much

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[lament stone]

how do i fill in the blanks

i cant seem to figure it out

[dull thorn]

Maybe develop the term (a+B) squared

U will sée something

[lament stone]

wdym delevope the term

develop

oops

[wraith arch]

distribute

[lament stone]

oh

[wraith arch]

or FOIL

[lament stone]

(a+b)^2 is a^2 + 2ab + b^2

now what

[whole coral]

What do you need to subtract from that to get a^2 + b^2?

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[lament stone]

ohh

-2ab

wait so i write (-2ab) right

cuz theres a minus already

[whole coral]

just 2ab, the - takes account of that

[lament stone]

okk

[whole coral]

Cause (a + b)^2 - 2ab = a^2 + b^2 is what they want from you, the 2ab being what to fill in

[lament stone]

tysm

[lament stone]

could u help me iwth 7?

the question below the one u jsut helped me solve

i get that its basically the same thing but (a+b)^2 has two positive signs and (a-b)^2 only has one

so what do i add to make them equal

[whole coral]

Well, expand them both, and see what you need to add to (a - b)^2 to get (a + b)^2

[lament stone]

i need -2ab to be +2ab

but idk how to do tht

[whole coral]

So, -2ab + [what] is equal to +2ab?

[lament stone]

ohh

okk i got it

tysm

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[vale obsidian]

im asked to find the probability that there is at most two defective lamps in a random sample of 30 lamps, given that there is at least 1 defective. The probability that a lamp is defective is 0.05.

[vale obsidian]

So i should just do binomial cd on my calculator where x = 1 and numtrials = 29 right? then i subtract this from 1

[frail yarrow]

Not really

[vale obsidian]

why?

[frail yarrow]

Do you recall how to calculate probability given a certain event?

[vale obsidian]

bayes theorem

yes

ill try that

[frail yarrow]

I feel like you're overcomplicating th ings

[vale obsidian]

yeah i think so

kind of too much

[frail yarrow]

Bayes is only useful if you already have P(B given A)

[vale obsidian]

how would i calculate P(A n B)

like binomial pd 1 + binomial pd 2?

[frail yarrow]

Nope

What do you think event A and event B are?

[somber onyx]

Pretty sure yes

[vale obsidian]

Event A = There are at most 2 deffective lamps
Event B = There is at least 1 deffective lamp

[somber onyx]

$$P(X \le 2|X \ge 1)$$
is basically
$$P(1 \le X \le 2)$$

[solid kiln]

Akuma

[vale obsidian]

hm

but is there an alternative way i should be thinking about this?

[somber onyx]

You're on the right track here

[vale obsidian]

then i divide by the prob of event B

[somber onyx]

For P(X>=1) you just find the complement

[vale obsidian]

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[whole coral]

Characteristic 3

a^3 = 0

1 + a can you find a way to use those two above facts, and find some expression involving 1 + a?

(also notice the ring is commutative!)

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[stoic flame]

Thoughts on this?

(1+a)^3 = a^3 + 3a^2 + 3a + 1 = 1

Not sure when we use commutativity

[whole coral]

Exactly that! Commutativity is what allows you to do that expansion

[stoic flame]

How come that lets us expand it?

[whole coral]

Namely that for any commutative ring you can e.g. binomially expand

[wooden plover]

do you even need commutativity here yeah

it's just powers of a

[whole coral]

(actually just noticed this, cause it's 1 )

[wooden plover]

oh well

if it makes ppl less stressed why not add it

[whole coral]

Was thinking of a different problem where it was similar but you needed the commutativity, showing the sum of nilpotent elements is nilpotent

[whole coral]

[in general though, that allows you in thinking of (a + b)^k to swap the order of multiplication in the products you end up finding, which was the thing I was thinking of haha]

[marble wharf]

I mean you dont even need char 3, do you? just do geometric series on 1+a^3

[wooden plover]

yeah indeed

they just wanted to have fun with (1+a)^3 = 1^3 + a^3 here ig

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@wraith hinge Has your question been resolved?

[pulsar plaza]

!status

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3. I got an answer but I was told that it's wrong.
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5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[pulsar plaza]

what have u done so far?

ok

then think about it this way

the fox is ahead 72 jumps

so when the dog starts it will be at 0 jumps

distance wise (as you said) the fox is ahead 3x72 = 216 distance ahead

every jump the dog gains 8 distance

whereas as the fox gains 3 distance to its already 216 distance

so can you create an algebraic relationship

for each jump j what dist_d and dist_f equals ?

thats a number

[pulsar plaza]

im looking for this

no total distance

dist_f = 3j + 216

dist_d = 8j

how far they have travelled

they as in each of them

i gtg now sry

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[chilly raven]

https://cdn.discordapp.com/attachments/1161035250676285512/1231424514072514610/IMG_1684.jpg?ex=6636e872&is=66247372&hm=7117432ea83e8174990da52d4447acce58ca9d00e4abc976c5088e4abbf96991&

[chilly raven]

I tried Ratio test

But I get L = 0

Nvm I'm stupid

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[obtuse oyster]

I am just confused on what the question is asking, and how would I go about figuring said question out

[obtuse oyster]

Is it just asking me to put the values provided into the answer boxes?

Like, A-B would be 125 < 20?

Is it that simple? or am I overcomplicating it

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[jolly kayak]

so I believe the GCD's imply that a,b and c are all coprime and thus only share the divisor of 1

[normal bronze]

Is tau the lcm

[jolly kayak]

the tau function is the amount of divisors for some n

ex: tau(12) = 6 as 12 has the divisors 1, 2, 3, 4, 6, 12

[normal bronze]

Right right

[jolly kayak]

therefore I think tau(abc) should simply be tau(a) + tau(b) + tau(c) - 2 (excluding that they all share a divisor of 1)?

[deep niche]

Then it's not the sum, but the count

[normal bronze]

I believe that’s correct yes

They are all coprime

[jolly kayak]

oopsie yeah that would be sigma haha

is there a more rigorous way to prove this?

[deep niche]

Me too

[normal bronze]

Prime expansion

[normal bronze]

.

[jolly kayak]

gotcha

[normal bronze]

Actually that seems incorrect

You create new divisors

For example if you take 2 and 3

[jolly kayak]

tau(2) = 2, tau(3) = 2

tau(6) = 4 no?

[normal bronze]

Yes but you need to remove 1

From what you said

[jolly kayak]

right

because of the shared 1

[deep niche]

Oh right, forgot about the number ab

[normal bronze]

Yeah so you need to account for the divisors being multiplied

[deep niche]

So you need to add 2^n - n

[jolly kayak]

I recnetly proved this

does it help?

[deep niche]

Oh right, the divisors create new factors too

[deep niche]

In that case this extends to the ternary case

[vagrant marsh]

youre essentially proving the identity extends to more pairwise coprime numbers

[deep niche]

So it would be T(a)T(b)T(c)

[jolly kayak]

so t(abc) = t(a)*t(b)*t(c)

[vagrant marsh]

yes

[jolly kayak]

hmm, that is shockingly similiar to this then huh

[vagrant marsh]

as long as a,b,c are pairwise coprime

suggestion: try proving gcd(a,bc)=1

then proceex

[jolly kayak]

do I need it?

like couldn't I just write the prime factorization of a,b, c

[vagrant marsh]

sure

it was a suggestion after all

[jolly kayak]

and then say tau(a), tau(b), tau(c) have no common factors

and thus tau(a) * tau(b) * tau(c) = tau(a * b * c)?

[vagrant marsh]

gcd(tau(a),tau(b)) can be greater than 1 despite gcd(a,b)=1

take 3 and 5 for example

[deep niche]

There are many similar things in math. Essentially moving from a pairwise statement to an arbitrary number of values.

[jolly kayak]

wait did I mess up the proof then?

sorry just trying to understand

[vagrant marsh]

i would say follow my suggestion

[deep niche]

You mean that any Pi is distinct from any Qi

[jolly kayak]

ah yeah that's probably better wording

what I meant is like no (ki + 1) term will be shared

[jolly kayak]

I think I have it

but why is this mission critical?

[jolly kayak]

i guess now it's basicallly equiavelent to this proof but why make the distinction?

[trim jolt]

@jolly kayak Has your question been resolved?

[deep niche]

You're trying to apply the pairwise proof to 3 (or more) terms. Before that can be done, you must first establish, as GBF said, that a and bc satisfy the condition of gcd(a, bc) = 1.

I.e. That the pairwise property holds "associatively"

[jolly kayak]

ah makes sense

can I post the full proof I have here in ~ 10 minutes to check for accuracy

I can do a scanned copy as well if that's easier

question 3 for reference

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[jolly kayak]

@deep niche

did I cook my bruda

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@jolly kayak Has your question been resolved?

[deep niche]

Looks good 🙂

[jolly kayak]

.close

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[amber saffron]

confused on how to find the inverse

[amber saffron]

.close

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[olive nymph]

.reopen

[wispy raptor]

how to prove a limit does not exist using epsilon delta?

[dull thorn]

What's the exercise

[wispy raptor]

like $\lim_{x\to 0} \frac1x$ for example

[solid kiln]

babario

[dull thorn]

Yeah maybe with the definition and the continuity of this function

[wispy raptor]

How with the definition tho?

Proving something exists is pretty obvious, i just dont understand how to properly prove something does not exist?

[dull thorn]

Ok so I think in this type of problem WE Can go for the absurd thus WE Can use our definition

Like suppose the limit exist

So what do we know with the definition

And with the unicity of the limit you go to a contradiction

I think it will go something like that

[wispy raptor]

So suppose $0 < |x| < \delta \implies \left|\frac1x - L\right| < \varepsilon$?

[solid kiln]

babario

[dull thorn]

What is delta and epsilon and L

[wispy raptor]

$L\in \Bbb{R}$?

[solid kiln]

babario

[wispy raptor]

But since its a contradiction we should show that for all L there exists epsilon that for all delta exists x where this doesnt hold true right?

[dull thorn]

L IS your limit right

[wispy raptor]

Yea

[dull thorn]

Ok maybe I will do something like it exists a sequence in R*+ such that it converges to 0+ take 1/n and f(1/n) converges to plus infinity with f(x) =1/x

Thus the limit in 0+ is plus infinity

Make thé same with another sequence in 0- like -1/n and u will have that the limit IS -infinity so with this 2 limits no equal it doesn't have a limit on 0

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[wispy raptor]

Yea that works but im curious for specifically using epsilon delta to proof

[dull thorn]

Ok I will try to do it so let me some time

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[subtle delta]

express (-1+i)^5 in polar form

[shy nacelle]

Hi

[cosmic carbon]

can you write -1+i in polar form

[dusty gate]

you can first convert (-1 + i) into polar form

[subtle delta]

yo dude

[shy nacelle]

Woah what you did.

[subtle delta]

de moivre theorem

[subtle delta]

i did

[shy nacelle]

Bro good

Cos 3pi/4 + isin 3pi/4

• 4√2

[subtle delta]

this is polar form

[shy nacelle]

In polar form itt will.be r e^itheta

R is modulus value

[subtle delta]

no i dont follow that one

it's too hard

[dusty gate]

???

[subtle delta]

i havent studied that one yet

[dusty gate]

why

[shy nacelle]

You said you want jn polar form

[subtle delta]

look

[dusty gate]

^

[subtle delta]

i only kno the above one

[dusty gate]

they're both the same thing

anyways so

[subtle delta]

dk what the 2nd one even means

[dusty gate]

u gotta take the modulus

by 360 degrees

or 2pi

whichever one you're doing

[shy nacelle]

Yeah just thin 4√2 is outside

[subtle delta]

modulus by 360 degrees

??

[dusty gate]

,, \sqrt 2 (\cos (135\ang) + \sin(135\ang))

[solid kiln]

inf1425
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dusty gate]

right

[subtle delta]

exactly

[subtle delta]

.

[subtle delta]

you didnt put the iota

[dusty gate]

now u raise it to the power n by using de moivre's formula

[subtle delta]

do i have to rite the angle = 5(315) = 1575

or can i just rite 315

cuz both give equal values

[dusty gate]

its 135

[subtle delta]

no when i raise the power

[dusty gate]

for polar form you always express it as 135

and angles that are below the x axis are negative

not > 180

[subtle delta]

and after i use the de moivre theorem

what angkle am i gonna rite

315 or 1575

[dusty gate]

how are you getting 315?

or 1575

[subtle delta]

tan inverse of -1 is 315

[dusty gate]

its -1 + i not 1 - i

[subtle delta]

i didnt rite -45

[subtle delta]

same thing

[dusty gate]

?

[dusty gate]

can u plot (-1 + i) on the complex plane and see what angle you get from the x axis if drawing a line from the origin to (-1 + i)

[dusty gate]

yes

what angle is that

[subtle delta]

ummm

[dusty gate]

would it be 315

[subtle delta]

1135

135

[dusty gate]

yeah

[subtle delta]

180-45 = 135

[dusty gate]

135

and to the power 5 so the angle will be multiplied by 5

135 * 5 = 675

[subtle delta]

duhhhh

im a dummy

i wanted to say 675

[dusty gate]

modulo 360 u get 315 but you dont write 315

[subtle delta]

the hole time

[dusty gate]

u write -45

[subtle delta]

instead of 1575

[subtle delta]

hmmm

y do i dont rite that

[dusty gate]

for polar form u never write angles above 180

[subtle delta]

ok

[dusty gate]

its just a rule

if the angle is below the x axis it becomes negative and is taken clockwise

rather than anticlockwise

your final answer should be 4 sqrt(2) * (cos(-45) + isin(-45))

[subtle delta]

can i put the -45 out of the sine

[dusty gate]

as in

oh

you can but it won't have any use

polar form is normally written as r.e^(itheta)

[subtle delta]

just looking better lol

this one

[dusty gate]

,, 4\sqrt 2 e^{i \cdot -45^\circ}

[solid kiln]

inf1425

[dusty gate]

is what you would normally write

[subtle delta]

tf is this

[dusty gate]

so its no use taking the minux sign out

,, 4\sqrt 2 e^{-i \frac{\pi}{4}}

[solid kiln]

inf1425

[dusty gate]

to be more precise

but yeah so like

its no use takign it out

u could

[subtle delta]

damn that is looking like the arhenius equation

@dusty gate thanks!!!

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[lone mauve]

did I solve this correctly?

first i used cross product of v1 x v2

then to make 0 vector, i used back substitution for a, b, c

to make each entry in the vector = 0

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[daring dragon]

How do i do #36?

[solid kiln]

uoǝu

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[steep grove]

Hi, I don't rly understand what is happening in the second equal sign?

[steep grove]

(for context this is aproximation of differential equations using series)

[true sierra]

It's a simple multiplication really

Multiply the content of the brackets with the sum one by one

x multiplied by the sum gives the same sum, only that each member is now k+1 degree

Since x^k * x = x^k+1

and then the other member is 1 so it's just the sum again

[steep grove]

oh jezus christ

I was looking for some complicated series expansions

thanks hahaha

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[dim latch]

Can someone help me on number 8 I think its letter B but im not sure

[nimble stone]

,rccw

[nimble stone]

certainly cant be b

why do you think it is?

[dim latch]

Umm because its intersecting in -2

[nimble stone]

thats not how x intercepts work

y=2x doesnt hit the x axis at x=2

does it

[dim latch]

How does it work, I didn't really pay attention to that in class

[nimble stone]

do you understand the terms slope/gradient and y-intercept

[dim latch]

No

[dim latch]

No lol

[nimble stone]

the slope is how much y changes by when x increases by 1, the y intercept is where the line crosses the y axis, ie when x=0

straight lines have the form y=mx+c where c is the y intercept and m is the slope

you said m was -2, but that would mean the line was going down as x increases, not up

and you were claiming the y intercept was -4, which is also not true

[dim latch]

Oh ok then is it a?

[nimble stone]

si

[dim latch]

Thank you

You think you could help me with the other ones please?

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[visual hollow]

Explain why R[x], the ring of real polynomials in the variable x, has no
invertible elements which are not constant.

[visual hollow]

can i please get help on how to answer this, pls? (above, 6 and below, 4)

and also parts (i), (ii) and (iv) of this

for 4(ii), i've said it's not isomorphic because there doesn't exist any a,b in 2Z which will add up to give any of the odd numbers in Z

is that right??

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@visual hollow Has your question been resolved?

[lean prawn]

Not constant = degree at least 1

Multiplying a polynomial by some other nonzero polynomial never makes its degree smaller (true for constant polynomials, true for monomials, true for all nonzero polynomials by induction)

Therefore, for any non-constant polynomial, i.e. one with degree at least 1, there can exist no polynomial such that their product would be 1 (which has degree 0)

[visual hollow]

uh

(true for constant polynomials, true for monomials, true for all nonzero polynomials by induction)
is there supposed to be an "except" before this?

[lean prawn]

x^3 has degree 3

If you multiply it by any nonzero polynomial, it will have degree at least 3

I guess probably the best way to see that is to consider the cofficient of the highest degree term of the polynomial

so if you multiply it by 5x^10 + 2x^3 for example

Then the resulting polynomial will have a nonzero term of degree 13, which is at least 3

And I guess more generally if you have any arbitrary polynomial like 10x^3 + 2x^2 + 3 then if you multiply that by any polynomial of degree k where the highest degree term has the nonzero coefficient a, then the product polynomial will have a term of degree k + 3 with the coefficient 10a, which in particular is nonzero

and all of this means that you can't invert the expression because the constant polynomial P(x) = 1 has degree 0

[visual hollow]

(still here, just reading and understanding slowly)

and eating custard creams

[lean prawn]

You might want to remind yourself of how polynomials are multiplied together

[visual hollow]

why??

Multiplying a polynomial by some other nonzero polynomial never makes its degree smaller
i understand and agree with this

[lean prawn]

if you agree with that then that's pretty much all you need

[visual hollow]

just not the bracketed part

[lean prawn]

well I actually take back the part in parenthesis

[lean prawn]

I think this is a better way to see it

but the problem statement didn't really seem to ask for a proof anyway

so if you're convinced by "multiplying a polynomial by a nonzero polynomial never makes its degree smaller" then you can just roll with that

[visual hollow]

what's a monomial? 'ax^1'?

where a is real

[lean prawn]

a monomial is a polynomial that only consists of one nonzero term

So 1, x, x^2, 3x^3, are all examples of monomials

[visual hollow]

ohh

so powers don't matter

cool

[lean prawn]

yeah

a monomial only restricts the amount of summands, nothing else

[visual hollow]

what's a summand

[lean prawn]

a fancy term for a thing that's being summed

[visual hollow]

the things you're summing with

[lean prawn]

ye

[visual hollow]

ok

[lean prawn]

(but that's me speaking very loosely because I would certainly count 3x+5x as a monomial because it simplifies to 8x)

[visual hollow]

Therefore, for any non-constant polynomial, i.e. one with degree at least 1, there can exist no polynomial such that their product would be 1 (which has degree 0)

i don't get this - there can exist no polynomial such that their product would be 1 (which has degree 0)

[visual hollow]

yh i would too

[lean prawn]

So you're given a polynomial of degree at least 1 (a non-constant polynomial)

If you can find some other (nonzero) polynomial such that their product is 1 then you call that polynomial an invertible element

If no such polynomial exists, then the polynomial is not invertible

And my claim is that a polynomial like 5x^2 will have no other polynomial such that their product is 1

because if such a polynomial existed then multiplying 5x^2 by that polynomial would make the degree smaller

which we know is impossible

[visual hollow]

"their product"???? the non-constant degree 1 polynomial * what??

[lean prawn]

You're given 5x^2

If there exists some nonzero P such that 5x^2 * P = 1 then you call 5x^2 an invertible element of the ring

No such P, however, can exist, because if it did, then you'd use multiplication to lower the degree of 5x^2, which is impossible

[visual hollow]

ah ok

i get it

why 1 btw?

why do they have to multiply to 1

is it coz that's the multiplicative identity?

so if you multiply it by 5x^10 + 2x^3 for example
Then the resulting polynomial will have a nonzero term of degree 13, which is at least 3
x^3 * (5x^10 + 2x^3), is this what you mean?

[lean prawn]

Yes exactly

[lean prawn]

yup

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[fading pebble]

I have a model with three variables a_1, a_2 and w and I would like to know how I can find the optimal values of these three to minimize the error defined by the picture.

[fading pebble]

How would I go about doing that?

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[native nacelle]

Where should i start with

[vivid juniper]

have you tried partial fractions?

[swift veldt]

^

[native nacelle]

I didnt learn that

[swift veldt]

what did u learn

[native nacelle]

Is there any other way or show me how to do it

Here is what i did for the last problem

[swift veldt]

this looks like a weird way to do partial

from how i normally do it at least

but that looks like partials

[native nacelle]

Can u show me how u do it

[round mango]

take x^2 common from the denominator and let 1/x=t

[round mango]

partial fractions is just a methodological way of doing what you've done in the previous problem, except breaking 8 in terms of x^2 and x-2 is not easy mentally, so we form a system of equations by letting the fraction equal to A/x + B/x^2 + C/x-2.

[native nacelle]

I get

[round mango]

tho i try to avoid that as much as possible since forming that system of eqns can take a lot of time

[native nacelle]

$\int{\frac{2-4t}{1+2t}}dt$

[solid kiln]

San

[round mango]

no you messed up some signs

you should get 2+4t/2t-1

[native nacelle]

Wait ill go home first now at school

[round mango]

unless you put -1/x=t instead

which is also fine as long as you substitute it back correctly

[native nacelle]

I should get 2t+2ln|2t-1| right

[round mango]

where t=1/x yes

[native nacelle]

I need to write it down first

@round mango

[round mango]

perfect

[native nacelle]

Man tysm t=1/x was a life saver man

My teacher did partial fraction i cannot understand shi so many stuff going on

[round mango]

its fine, tho i recommend you to learn partial fractions as well, theyre handy(sometimes)

[native nacelle]

U would use partial in this?

[round mango]

actually what you've done here is partial fractions only, but without making the 'formal' equations

we ve basically broken down this thing into partial fractions

this one is mentally doable

but the other one with the cubic in the bottom was a little complex for breaking it mentally, so we need to form some equations

[native nacelle]

Idk what was it i did it like my extra class teacher

[native nacelle]

It sure is i cant understand it and i tried partial also but it was so complicated

[round mango]

Yeah it would have been calculative

doing stuff like that is my last resort

when i cant think of anything

[round mango]

do you want me to guide you through that method as well or are you good?

[native nacelle]

Nah im good ill watch some youtube later maybe

Tysm I couldn’t even think of t=1/x

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[dapper prairie]

What type of equation is this?

[grizzled ridge]

i cant help but you may want to translate that for someone who can help

[dapper prairie]

That line means consider the equation

They call it the equation

but I have never seen it before

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[gusty quarry]

Can someone look over this? So would something like.

If W= {v1,v2,v3,vn } is the set of vectors in the third quadrant vector space V, then W will be the basis if:
W spans W
W is linearly independent

[gusty quarry]

Make sense for number seven

[whole coral]

Yep, linearly independent spanning set would do as the definition

[gusty quarry]

Alright glad that answer works

Thank you

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[gaunt osprey]

I’m stuck on this question
10b, I have worked out 10a
I’m not sure how to workout the the constant and where to start.

[gaunt osprey]

@help

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@gaunt osprey Has your question been resolved?

[gaunt osprey]

<@&286206848099549185>

[earnest granite]

so we know that : (a-b)^5 = a^5 - 5a^4b + 10a^3b^2 - 10a^2b^3 + 5ab^4 - b^5

substituting for 1 and 4x we get :

[gaunt osprey]

X is 0.25?

[earnest granite]

(1-4x)^5 = 1 - 5(4x) + 10(4x)^2 - 10(4x)^3 + 5(4x)^4 - (4x)^5

[gaunt osprey]

Is that part b?

[earnest granite]

isn;t it ?

[gaunt osprey]

I find this confusing to understand

[earnest granite]

yes it's b

[earnest granite]

well I can explain it to you in two ways

on is the analytic and other one the quick one

which one would you like

[gaunt osprey]

Which one would I use more of in tests?

[earnest granite]

I don't know tbh

[gaunt osprey]

Could you explain the one that is easier to understand and to do? Just in your opinion

[earnest granite]

well it is easier the analytic way so

what you can do is

Oh wait

[earnest granite]

do you understand that?

[gaunt osprey]

Yes, so in this case a would be 1 and b would be 4?

[earnest granite]

no b would be 4x

[gaunt osprey]

No sorry that’s what I meant

[earnest granite]

oh ok

yes it is like that

but just for simplicity instead of multiplying by one I just skipped that

[gaunt osprey]

And how do I compare the coefficient

[earnest granite]

you also need to multiply it with the (1+ax+bx^2) part

[gaunt osprey]

But it says first 3 terms of the expansion of (1-4x)^5

[earnest granite]

well that works too

because you only need these to get your result because for any higher powers the product would exceed the second power of x with is where your expression (1 - 23x - 22x^2) stops

just multiply the terms you found in part a with (1+ax+bx^2)

[gaunt osprey]

I expanded the thing

But the number has many a and b in it

[earnest granite]

ok

what did you find

[gaunt osprey]

Sorry it keeps going sideways

[earnest granite]

it's ok

ok so first of all what you will ulimatly do is compare the coefficients of the three first powers of x (0,1,2) with (1-23x-222x^2)

so rewrite youre expression but this time ignore any term with power highr that 2

[earnest granite]

brilliant, so now we need to express a and b as coefficients of the powers of x to do that we rewrite the result in descending powers of x and get x^2 as a common factor for the 2 powers of x and x as commpn factor for the 1 powers of x.

[earnest granite]

yes! so know you just compare your coefficients and you find a and b

[gaunt osprey]

Oh I got it

Thank you

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[regal hatch]

what's my mistake

[split chasm]

what makes you think you made a mistake

reaching that result just means that'll be true where defined

[regal hatch]

idk it sounds wrong

[regal hatch]

oh

i thought i would find x = something

[hexed shard]

this result basically means that the equation is true for all x

[split chasm]

and since those functions are defined in the given interval
the whole interval will be the solution set

[regal hatch]

ok, thank you

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[gusty mulch]

Do you guys think the right side is a correct graph

[gusty mulch]

roughly

graph of the derivative

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@gusty mulch Has your question been resolved?

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[paper willow]

how to calculate the limit of this sequence

[pseudo grotto]

Look at degrees of num / denom

[paper willow]

this shit is unsolvable

please help!!!

,rotate

[wraith hinge]

so we have to convert it into a proper fraction?

[pseudo grotto]

You could I suppose

But each of the two terms has larger degree in the num than the denom

[paper willow]

this diverges then?

[pseudo grotto]

So individually the terms are tending to infinity

and you're taking their sum

[pseudo grotto]

Yes

[paper willow]

so the limit doesnt exist

,rotate

[paper willow]

is this enough evidence that the sequence diverges and the limit doesnt exist?

did I messed something up?

[pseudo grotto]

Hmm

I suppose you should actually get a common denom and combine the two fractions into 1 first

[paper willow]

interesting

[pseudo grotto]

Then factor out n^2 from num and denom

[paper willow]

thanks

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[wraith hinge]

Can someone help me cros reference?

[wraith hinge]

this is what I got

can anyone help me check if this is correct?

[whole coral]

Should be |c| you have at the end

But that's fine other than that, now you want to consider what that limit becomes to conclude convergence/divergence

[wraith hinge]

u think i should show more work or is that fine

[whole coral]

I think that's fine, if anything maybe e.g. divide the numerator and denominator by n^2 to make clearer why that fraction has limit 1, but other than that (and the previously mentioned |c|), it's fine-

As per before, now state what you want |c| to be for convergence, and for divergence-

[wraith hinge]

bet

i have another question to cross reference is that ok

[whole coral]

Sure

[wraith hinge]

@whole coral

[whole coral]

Also missing out absolute value signs for the first one (the second they state you're greater than zero, so I'll let you live ) oh nvm, the first one they say that too

But yeah, they'll converge when the limit is < 1, diverge when the limit is > 1

[whole coral]

interesting they do that here as well, c > 0 they really should have made you check both positive and negative values

[wraith hinge]

yea the questions feel pretty similar

[whole coral]

You just need to check whether you converge when c = 1 now

[do you?]

[wraith hinge]

idk?

it didnt say to at least

[whole coral]

Remember that the ratio test tells you that (for this one) c > 1 means you must diverge and 0 < c < 1 tells you you must converge, but you have to find other ways to check convergence for c = 1

if you set c = 1, what series do you get? Is it a convergent one?

[wraith hinge]

Um idk tbh

im blanking rn

but i'd say no?

[whole coral]

what series tests do you know?

[wraith hinge]

root, ratio, integral, limit comparison, alternating and absolute convergence

[whole coral]

well, I mean the integral test may help here-

I will save you and tell you that the series $\sum_{n = 1}^{\infty} \frac1{n^2}$ does converge (it in fact converges to $\frac{\pi^2}6$ in fact, but yea)

[solid kiln]

@whole coral

[wraith hinge]

mmmm

omg idk why i was brain farting. bruh plugged 1 on top for c and did the test

[whole coral]

Using the integral test, you can find the "$p$" series tests ( who names these) which tells you $\sum_{n = 1}^\infty \frac1{n^p}$ converges when $p > 1$ and diverges when $p < 1$

[solid kiln]

@whole coral

[wraith hinge]

i see wym now ok

this is another one I had

i wanted to know if it was correct to grab the integral of dy/y^3

or should I have just got the anti derivative and had it be y^4/4 instead of -1/2y^2

@whole coral c

[whole coral]

You do want the integral of 1/y^3 wrt y, but also you didn't do the left hand side right...

Notice you also have that x by itself too...

[wraith hinge]

wait x where

[whole coral]

The very first term is x + ...

[wraith hinge]

oh yea thats my b. I could put x in the right denominator right?

[whole coral]

numerator

[wraith hinge]

so itd be like $-\int \frac{x}{4\sqrt{x^{2+1}}}dx$

[solid kiln]

Draco

[wraith hinge]

accidentally put the 1 in the wrong place but yea

and it'd be like $-\frac{1}{2y^2}=\int \frac{x}{4\sqrt{x^2+1}}dx$

[solid kiln]

Draco

[whole coral]

Yep basically, and that right integral isn't too bad ⛓️

[wraith hinge]

and then $-\frac{1}{2y^2}=-\frac{1}{4}ln\left|x+\sqrt{x^2+1}\right|+C$

[solid kiln]

Draco

[whole coral]

Well, you don't need a ln for that one

[wraith hinge]

extra negative by accident

[wraith hinge]

mmm ok ok, but correct so far?

[whole coral]

I don't think it is

For one, you can do that integral by substituting, set, say, u = x^2 + 1 and then you get a nice integral

[wraith hinge]

$-\frac{1}{2y^2}=\frac{1}{4}\sqrt{u}$

[solid kiln]

Draco

[whole coral]

Yep, you get $\frac1{2y^2} = \frac14\sqrt{x^2 + 1} + c$, and now, you just need to find the $c$

[solid kiln]

@whole coral

[whole coral]

(remembering to put the original variable back and all!)

[wraith hinge]

$-\frac{1}{2\cdot \left(2\right)^2}=\frac{1}{4}\sqrt{\left(0\right)^2+1}$

[solid kiln]

Draco

[wraith hinge]

well + c

$-\frac{1}{8}:=:\frac{1}{4}:+:c$

[solid kiln]

Draco

[wraith hinge]

$-\frac{1}{8}-\frac{1}{4}=:\frac{3}{8},::C=:\frac{3}{8}$

[solid kiln]

Draco

[wraith hinge]

@whole coral hows it lookin

[whole coral]

nice!

[wraith hinge]

so i got it right?

fye

Thats all I ahve for rn. There are some other problems but I havent done them yet

thank you

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[sharp gust]

What am I messing or missing when trying to calculate the critical points to then find abs max and min?

[sharp gust]

Interval is [-1,6]

I’m making basic mistakes as usual

[lilac cloud]

Isn’t the derivative of the t term 1

[sharp gust]

Oooh right I left out the 1

[lilac cloud]

And then I don’t know what you did on the second last step, could you explain

[sharp gust]

Ignore that last part

This is where I am stuck - I can’t remember what to do to get t= for the critical numbers to see what lay within the interval

[lilac cloud]

You definitely can’t take the ^2/3 away from the t like that